Problem: Lenmana started studying how the number of branches on her tree grows over time. The number of branches increases by a factor of $\dfrac{11}{6}$ every $3.5$ years, and can be modeled by a function, $N$, which depends on the amount of time, $t$ (in years). When Lenmana began the study, her tree had $48$ branches. Write a function that models the number of branches $t$ years since Lenmana began studying her tree. $N(t) = $
Explanation: The strategy We can model the situation with an exponential function of the general form A ⋅ B f ( t ) A\cdot B\^{ f(t)}, where $A$ is the initial quantity, $B$ is a factor by which the quantity is multiplied over constant time intervals, and $f(t)$ is an expression in terms of $t$ that determines those time intervals. Let's use the given information to determine $A$, $B$, and $f(t)$. Understanding what's given We are given that the initial number of branches is $48$, and the number of branches increases by a factor of $\dfrac {11}{6}$ in $3.5$ years. This means that the initial quantity is $A=48$ and the factor is $B=\dfrac{11}{6}$. We need to find $f(t)$ based on the fact that the quantity is multiplied by $\dfrac{11}{6}$ every $3.5$ years. Finding the expression in the exponent We know that the number of branches is multiplied by $\dfrac{11}{6}$ every $3.5$ years. This means that each time $t$ increases by $3.5$, $f(t)$ increases by $1$. Therefore, $f(t)$ is a linear function whose slope is $\dfrac{1}{3.5}$. When the initial measurement is made, the number of branches hasn't changed. So $N(0) = 48$, which means that $f(0)=0$. [Why?] Therefore, $f(t)$ must be $\dfrac{t}{3.5}$. Summary We found that the following function models the number of branches $t$ years since Lenmana began studying her tree. N ( t ) = 48 ⋅ ( 11 6 ) t 3.5 N(t)=48\cdot \left(\dfrac{11}{6}\right)\^{ \frac{t}{3.5}}